Question & Answer
QUESTION

Find whether the following series is convergent or divergent:
\[1 + a + \dfrac{{a(a + 1)}}{{1.2}} + \dfrac{{a(a + 1)(a + 2)}}{{1.2.3}} + .....\]

ANSWER Verified Verified
Hint: The binomial expansion of \[{(1 - x)^{ - n}}\] is given as \[{(1 - x)^{ - n}} = 1 + nx + \dfrac{{n(n + 1)}}{{2!}}{x^2} + \dfrac{{n(n + 1)(n + 2)}}{{3!}}{x^3} + .....\]. Use this formula to find the value of the given sum of infinite series.

Complete step by step answer:
The binomial theorem or the binomial expansion is a result of expanding the powers of binomials or the sum of two terms. The coefficients of the terms in the expansion are called the binomial coefficients.

The binomial expansion of \[{(1 - x)^{ - n}}\] is given as follows:

\[{(1 - x)^{ - n}} = 1 + nx + \dfrac{{n(n + 1)}}{{2!}}{x^2} + \dfrac{{n(n + 1)(n + 2)}}{{3!}}{x^3} + .....\]

It can also be proved using the Taylor series.

Let \[f(x) = {(1 - x)^{ - n}}\], then, we have:

\[f(0) = 1\]

The first derivative of f is given as follows:

\[f'(x) = n{(1 - x)^{ - n - 1}}\]

The value of this function at x = 0 is given as follows:

\[f'(0) = n\]

The second derivative of f is given as follows:

\[f''(x) = n(n + 1){(1 - x)^{ - n - 2}}\]

The value of this function at x = 0 is given as follows:

\[f''(0) = n(n + 1)\]

The Taylor series expansion of a function at x = 0 is given as follows:

\[f(x) = f(0) + \dfrac{{f'(0)}}{{1!}}x + \dfrac{{f''(0)}}{{2!}}{x^2} + ...\]

Then, the value of \[{(1 - x)^{ - n}}\] is given as:

\[{(1 - x)^{ - n}} = 1 + nx + \dfrac{{n(n + 1)}}{{2!}}{x^2} + \dfrac{{n(n + 1)(n + 2)}}{{3!}}{x^3} + .....\]


We substitute the value of n as a, then, we have:

\[{(1 - x)^{ - a}} = 1 + ax + \dfrac{{a(a + 1)}}{{2!}}{x^2} + \dfrac{{a(a + 1)(a + 2)}}{{3!}}{x^3} + .....{\text{ }}...........{\text{(1)}}\]

The expression given in the question is the same as the expression in equation (1) for the value of x as 1.

The left-hand side of the equation for x = 1 becomes infinity for all positive values of a.
Hence, the given expression diverges for a > 0.

The function itself is defined only for positive values of a.

Hence, for a = 0, then the value of the expression becomes 1. Hence, the given expression converges for the value of a = 0.

In conclusion, the given expression is divergent for the values of a > 0.

Note: You can observe that for any positive value of a, the numerator is always greater than the denominator, hence each therm is greater than 1, and hence, the sum of the infinite series is divergent.